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(C) The cell reaction is based on the oxidation of $HSnO_2^{-}$ and reduction of $Bi_2O_3$.The cell potential $E_{cell} = E^{\circ}_{cell} - \frac{0.0

Vedclass · Accepted Answer

$78$

Vedclass · Answer

(C) The cell reaction is based on the oxidation of $HSnO_2^{-}$ and reduction of $Bi_2O_3$.The cell potential $E_{cell} = E^{\circ}_{cell} - \frac{0.059}{6} \log Q$.$E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode} = -0.44 - (-0.9) = 0.46 \ V$.The reaction is $3 \ HSnO_2^{-} + Bi_2O_3 + 3 \ OH^{-} + 3 \ H_2O \rightarrow 3 \ Sn(OH)_6^{2-} + 2 \ Bi$.$Q = \frac{[Sn(OH)_6^{2-}]^3}{[HSnO_2^{-}]^3 [OH^{-}]^3} = \frac{(0.5)^3}{(0.05)^3 [OH^{-}]^3} = \frac{1000}{[OH^{-}]^3}$.$0.2353 = 0.46 - \frac{0.059}{6} \log \frac{1000}{[OH^{-}]^3} = 0.46 - \frac{0.059}{2} \log \frac{10}{[OH^{-}]}$.$0.2247 = 0.0295 \log \frac{10}{[OH^{-}]} \implies \log \frac{10}{[OH^{-}]} = 7.6$.$1 - \log [OH^{-}] = 7.6 \implies pOH = 6.6$.$pH = 14 - 6.6 = 7.4$.Using Henderson-Hasselbalch equation: $pH = pKa + \log \frac{[HCO_3^{-}]}{[H_2CO_3]}$.$7.4 = 6.11 + \log \frac{5x / (x+10)}{20 / (x+10)} = 6.11 + \log \frac{5x}{20}$.$1.29 = \log \frac{x}{4} \implies \frac{x}{4} = 19.5 \implies x = 78$.

Given :	$E^{\circ}_{HSnO_2^{-} \mid Sn(OH)_6^{2-}} = -0.9 \ V$
	$E^{\circ}_{Bi_2O_3 \mid Bi} = -0.44 \ V$
	$pKa_{(H_2CO_3)} = 6.11$
	$\frac{2.303 \ RT}{F} = 0.059 \ V$
	$Antilog(1.29) = 19.5$

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